I have read that to measure the position of a particle to an accuracy $\Delta x$, we need light of wavelength $\lambda < \Delta x$. Is it true? Why is it so?

# [Physics] Why must the wavelength of light be smaller than uncertainty in position

electromagnetic-radiationquantum mechanicswaves

#### Related Solutions

It wouldn't work simply because there doesn't exist any entangled – or unentangled – state that would violate the uncertainty principle. The inequality $$\Delta x_1\cdot \Delta p_1 \geq \frac\hbar 2$$ is a universal law of physics (which may be easily proved as a mathematical theorem if one assumes that the quantities are described by the maths of quantum mechanics) and it holds regardless of values or uncertainty or properties of $x_2,p_2$ (properties of another particle) or their entanglement with $x_1,p_1$ (subscripts $1,2$ label the two particles here).

So you can prepare a state in which $x_1$ is sharply determined and $p_2$ is sharply determined – they commute with each other – but because $x_1$ is sharply determined, it follows that $p_1$ is completely uncertain and similarly $x_2$ is completely uncertain, so the two particles just can't have the same positions or the same momenta. They're not entangled in the usual sense.

Alternatively, you may prepare entangled states in which the two particles always have the same position $$ |\psi\rangle = \int_{-\infty}^{+\infty} dx \, f(x)\ |x\rangle \otimes |x+a\rangle $$ where the first tensor factor describes the first particle and the second tensor factor describes the other particle (I added a variable shift $a$ so that the particles don't overlap if you don't want to). But for a particle in this state, we can prove that the uncertainty $\Delta p_2$ is equal to the uncertainty $\Delta p_1$ simply because the state is totally symmetric with respect to the exchange of the two particles so the calculations of $\Delta p_2$ and $\Delta p_1$ are manifestly the same and yield the same results.

Because $\Delta p_1\geq \hbar / 2(\Delta x_1)$ by the uncertainty principle, it follows that – in this state – $$\Delta p_2 \geq \frac{\hbar}{2\cdot \Delta x_1} $$ as well. The momentum of the second particle is greater than the usual multiple of the inverse uncertainty of the position of the first particle. This inequality doesn't hold for a general state because the two particles have nothing to do with each other. But if you impose an additional condition that the particles are entangled, then their properties are really the same and one may prove additional inequalities.

Your problem is that you seem to think that you are free to impose an arbitrary collection of conditions on the state vector and it still exists. But you're not free to do so. In other words, if you do so, your condition have no solution. Such states can't exist in Nature. They're mathematically impossible.

In Bohr's theory the smallest possible orbital angular momentum is $\hbar$. The measured value is $0$. On the other hand the picture developed by solving the (time independent) Schödinger equation reproduces the energy levels from Bohr's model *and* gets the minimum angular momentum and the angular momentum step size right (it also fives you the quantization of the projections of the angular momentum). Add Pauli exclusion to the Schroödinger picture and you can get the shell filling rules and explain why the periodic table has the structure that it does which is another thing that Bohr's atom couldn't do correctly.

Bohr is out because it makes incorrect prediction.

## Best Answer

It is because the wavelength is being used as a yardstick to measure the object ( actually the uncertainty in its position). You can measure a 10 meter object with a meter stick, but you can't measure a 10 cm object with the meter stick (assuming no markings on the stick)